$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$ $$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$$
I tried this 
$\frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1} = 1-2\sqrt{\frac{\arctan(x)-\frac{\pi}{4}}{x-1}}.\frac{1}{\sqrt{x-1}}$
I am stuck here ! 
 A: Let's compute
$$
\lim_{x\to1^+}\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}
$$
One way could be to substitute
$$
\sqrt{\arctan x-\frac{\pi}{4}}=t
$$
so
$$
\arctan x=t^2+\frac{\pi}{4}
$$
and therefore
$$
x=\tan(t^2+\tfrac{\pi}{4})=\frac{\tan(t^2)+1}{1-\tan(t^2)},
\qquad
x-1=\frac{2\tan(t^2)}{1-\tan(t^2)}
$$
so we get
$$
\lim_{t\to0^+}\frac{t(1-\tan(t^2))}{2\tan(t^2)}=
\lim_{t\to0^+}\frac{1}{2t}\frac{t^2}{\tan(t^2)}(1-\tan(t^2))
$$
The first factor goes to $\infty$, the other two go to $1$, so the overall limit is $\infty$.
Now
$$
\lim_{x \to 1^+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}=
\lim_{x\to1^+}\biggl(1-\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}\,\biggr)
=-\infty
$$
A different way is considering that
$$
\lim_{x\to1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1}
$$
is the derivative of $\arctan$ at $1$, so it is finite (precisely $1/2$); then our limit is
$$
\lim_{x\to 1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1}
\frac{1}{\sqrt{\arctan x-\frac{\pi}{4}}}=\infty
$$
A: $$
\begin{aligned}
\lim_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}
& = \lim _{t\to 0^+}\left(\frac{t+1-\sqrt{\arctan \left(t+1\right)-\frac{\pi }{4}}-1}{t+1-1}\right)
\\& = \lim _{t\to 0^+}\left(\frac{t-\sqrt{\arctan \left(t+1\right)-\frac{\pi \:}{4}}}{t}\right)
\\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{4\arctan \left(t+1\right)-\pi \:}}{2t}\right)
\\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{\pi +2t-t^2+o\left(t^2\right)-\pi \:\:}}{2t}\right)
\\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{-t^2+2t}}{2t}\right)
\\& =\frac{1}{2} \lim _{t\to 0+}\left(2-\sqrt{\frac{2}{t}-1}\right)
\\& = \color{red}{-\infty}
\end{aligned}
$$
Solved with substitution $t = x-1$ and Taylor expansion
